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Let $\Sigma_2$ denote the genus two surface. ( i.e., the connected sum of two tori ). I need to compute the homology of $\Sigma_2$ with three disjoint holes on it's surface. Call this resulting space $X$. I also need to compute $H^*(X,\partial X)$.

I wasn't really sure how to approach this. If there were only 1 hold I could deformation retract everything and obtain a wedge of circles, but here I can't do this. I thus tried approaching it using Mayer-Vietoris long exact sequence. In particular, with $A =$ 3 disjoint disks covering the holes and $B= X$, we have $A\cap B=$ the disjoint union of $3$ circles and $A\cup B = \Sigma_2$. I then computed the homology groups of $\Sigma_2$ using cellular homology and obtained $H_1(\Sigma_2) = \mathbb(Z)^4, H_0(\Sigma_2) =H_2(\Sigma_2)= \mathbb{Z}$. The portion of long exact sequence is then for $H_2(X)$: $$0 \rightarrow H_2(X) \rightarrow H_2(\Sigma_2)=\mathbb{Z} \xrightarrow{f} H_1(S_1)^3 = \mathbb{Z}^3...$$ $f$ must be injective, hence $H_2(X) = 0$. The part of the long exact sequence for $H_1(X)$ is: $$\mathbb{Z}^3 \rightarrow H_1(X) \xrightarrow{\psi} H_1(\Sigma_2)=\mathbb{Z}^4 \xrightarrow{\phi} H_0(S_1)^3 = \mathbb{Z}^3 $$ I am having trouble understanding what the map $\phi$ is here. I would like to understand how to find out what the relevant maps actually are here. I cannot seem to make more progress than this.

Rohan Nuckchady
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  • I believe it should be easier to proceed inductively. First, remove a point and calculate, Then, remove two points and calculate. By this point, you should be able to figure out the general case. – juan diego rojas Jun 08 '20 at 21:38
  • Alternatively, you can calculate the fundamental group and then take its abelianization. Van Kampen's theorem yields an easy way to do it. See https://math.stackexchange.com/questions/622864/the-fundamental-group-of-a-riemann-surface-with-n-points-removed – juan diego rojas Jun 08 '20 at 21:45

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You can deformation retract everything into a wedge sum of circles yet again.

Start by placing one hole to one copy of your torus, and then place two holes on your second copy of torus. Now insert a hole in each copy, with the intent to glue your two copies of tori along the boundary of those two discs at the end.

Overall, you have a copy of a torus with 2 holes, and a second copy with 3 holes. The first copy (which now has 2 holes) deformation retracts to a wedge sum of 3 circles. The second copy (which now has 3 holes) deformation retracts to a wedge sum of 4 circles. So you now have two disjoint bouquets of circles, one with 3 circles, the other one with 4 circles.

To perform the connected sum, you identify one of those circles (remember, you have removed a hole from each torus, with the intent to join those boundaries at the end).

Overall you have a wedge sum of $7-1=6$ circles, which has very easy homology.

To compute $H^*(X,\partial X)$, you can use Lefschetz duality.

Locally unskillful
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